1. **Problem:** Prove the set identity $$A \cap (B - C) = (A \cap B) - (A \cap C)$$. 2. **Recall definitions:** - Set difference: $$B - C = \{x \mid x \in B \text{ and } x \notin C\}$$. - Intersection: $$A \cap B = \{x \mid x \in A \text{ and } x \in B\}$$. 3. **Rewrite left side:** $$A \cap (B - C) = \{x \mid x \in A \text{ and } x \in B - C\} = \{x \mid x \in A, x \in B, \text{ and } x \notin C\}$$. 4. **Rewrite right side:** $$(A \cap B) - (A \cap C) = \{x \mid x \in A \cap B \text{ and } x \notin A \cap C\}$$ $$= \{x \mid x \in A, x \in B, \text{ and } x \notin A \text{ or } x \notin C\}$$ Since $x \in A$ is required, $x \notin A$ is false, so this reduces to: $$= \{x \mid x \in A, x \in B, \text{ and } x \notin C\}$$. 5. **Compare both sides:** Both sides describe the set of elements $x$ such that $x \in A$, $x \in B$, and $x \notin C$. 6. **Conclusion:** Therefore, $$A \cap (B - C) = (A \cap B) - (A \cap C)$$ is true. This completes the proof.